Physics 215 Quantum Mechanics 1 Assignment 1

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1 Physics 5 Quantum Mechanics Assignment Logan A. Morrison January 9, 06 Problem Prove via the dual correspondence definition that the hermitian conjugate of α β is β α. By definition, the hermitian conjugate of an operator A is defined by A n n A under the dual correspondence. Let n be and arbitrary vector. Then, we know that α β n β n α = n β α 3 = n β α 4 5 Since this hold for all n, we conclude that α β = β α 6

2 Logan A. Morrison Assignment Problem Page / 9 Problem Prove that, in the absence of degeneracy, a sufficient condition for the following to be true c b b a a b b c = c b b a a b b c 7 b b b where as usual a is and eigen-ket of A etc. is that [A, B] = 0 or that [B, C] = 0. Assume that both A and B are self-adjoint. Additionally, assume that [A, B] = 0. Then, AB a = BA a 8 = a B a 9 Therefore, B a is an eigenket of A with eigenvalue a. Since there is no degeneracy, there is only one eigenvector of A corresponding to the eigenvalue a : a. Therefore, B a is proportional to a. That is, B a = b a. In othere words, the eigenkets of B are proportional to the eigenkets of A. Thus, we can write b = γ a 0 Since the states are properly normalized, b b = and a a =, we require that γ =. Note that since the operators are self-adjoint, the eigenvectors are orthogonal. Therefore, b a = 0 unless b = γ b This yields: b c b b a a b b c = c γ a a γ a a γ a b γ c = γ 4 c a a c = c a a c Since B is hermitian, b b b b = b b = I 4 where I is the identity operator. b b c b b a a b b c = c I a a I c 5 Thus, the two expressions are equal and we have proved the equality in equation 7. = c a a c 6

3 Logan A. Morrison Assignment Problem 3 Page 3 / 9 Problem 3 Show that for the S z ; + state of a spin system S x S x = /4 First, from class, we know that We can invert these equations to obtain Using this, we find that S x, + = S z, + + S z, 7 S x, = S z, + S z, 8 S z, + = S x, + + S x, 9 0 S x = S z ; + S x S z ; + and = S x; + + S x ; S x S x S x ; + + S x ; = S x; + + S x ; S x S x; + S x; 3 = S x; + + S x ; 4 S x; S x; 4 = = 4 6 S x = S z ; + S x S z ; + 7 = S x; + + S x ; S x S x ; + + S x ; 8 = S x; + + S x ; S x; + S x; 9 = 30 = 0 3 and hence, S S x S x = /4. Physically, we know that σ S x = x S x = / where σ S x is the uncertainty of the value of S x for a particle in the state S x ; +. Since there is a probability of / that the value / will be measured and a probability of / that the value / will be measured.

4 Logan A. Morrison Assignment Problem 4 Page 4 / 9 Problem 4 A two-state system is characterized by the Hamiltonian H = H + H + H + 3 where H, H and H are real numbers with the dimension of energy, and and are eigenkets of some observable H. FInd the energy eigenkets and corresponding energy eigenvalues. To solve this problem, we should choose a representation for our states. Let = 0 33 and In this representation, we have that Now, let s find the eigenvalues: H = 0 = H H H H = deth λi 36 = H λ H H λ 37 H = H λ H λ H 38 = H H H + H λ + λ H 39 Therefore, λ = H + H ± H + H 4H H + 4H = H + H ± H H + 4H Thus, the two possible energies are: E + = H + H + H H + 4H 43 and E = H + H H H + 4H 44

5 Logan A. Morrison Assignment Problem 4 Page 5 / 9 Now let s find the energy eigenstate corresponding to E +. To do so, we will determine the null space of H E + I: Thus, 0 = H E + I E + H E = + H a H H E + b H H and therefore, choosing b = H H + 4H H 0 = a H H a = H H H H H H H H H + 4H a b H H + 4H + bh 48 H b 49 H H + 4H H H + 4H, we find H E + = H H H H + 4H 50 And now let s find the energy eigenstate corresponding to E. To do so, we will determine the null space of H E I: Thus, 0 = H E I E H E = H a H H E b H H + and therefore, choosing b = H H + 4H H 0 = a H H + a = H H H H + H H + H H H + 4H a b H H + 4H + bh 54 H b 55 H H + 4H H H + 4H, we find H E = H H + H H + 4H 56 Clearly these states are not normalized to. We could do so by defining E + = E + E + E + and a similar expression for E.

6 Logan A. Morrison Assignment Problem 5 Page 6 / 9 Problem 5 Construct the transformation matric that connects the S z diagonal basis to the S x diagonal basis and show that your result is consistent with the general relation U = b r a r. 57 r From class, we know that: S x ; + = S z ; + + S z ; 58 S x ; = S z ; + S z ; Thus, in the usual representation of S z ; + and S z ;, i.e. 0 S z ; + = and S 0 z ; = one can clearly see that the transformation matrix is: U = 6 6 In other words, S x ; + = U S z ; + 63 S x ; = U S z ; 64 In a representation-free form, U is: U = S z ; + S z ; + + S z ; S z ; + + S z ; + S z ; S z ; S z ; 65 Let s check that this transformation matrix yields the appropriate transformation: U S z ; + = S z ; + S z ; + + S z ; S z ; + + S z ; + S z ; S z ; S z ; S z ; + 66 = S z ; + S z ; + S z ; + + S z ; S z ; + S z ; + + S z ; + S z ; S z ; + S z ; S z ; S z ; + 67 = S z ; + + S z ; 68 = S x ; + 69

7 Logan A. Morrison Assignment Problem 5 Page 7 / 9 and U S z ; = S z ; + S z ; + + S z ; S z ; + + S z ; + S z ; S z ; S z ; S z ; 70 = S z ; + S z ; + S z ; + S z ; S z ; + S z ; + S z ; + S z ; S z ; S z ; S z ; S z ; 7 = S z ; + S z ; 7 = S x ; 73 Now, we can write our transformation matrix in a slightly different form: U = S z ; + S z ; + + S z ; S z ; + + S z ; + S z ; S z ; S z ; S z ; + 74 which is of the form r = S z ; + + S z ; S z ; + + S z ; + S z ; S z ; 75 = S x ; + S z ; + + S x ; S z ; 76 b r a r.

8 Logan A. Morrison Assignment Problem 6 Page 8 / 9 Problem 6 An operator A, corresponding to an observable α, has two normalized eigenstates φ and φ, with eigenvalues a and a. An operator B, corresponding to an observable β, has normalized eigenstates χ and χ. The eigenstates are related by φ = χ + 3 χ, φ = 3 χ χ. 77 α is measured and the value a is obtained. If β is then measured and then α again, find the probability of obtaining a a second time. First off, let s invert equation 78: χ = φ + 3 φ, χ = 3 φ φ. 78 Assume that α is measured and a is obtained. Then we know that the state of our system is φ. Now if we measure β and then α, there are two ways of obtaining a again. The first case is that we measure β and obtain χ and then measure α and obtain a. The second case is we measure β and obtain χ and then measure α and obtain a. The probability of obtaining a after the two measurements will be The probability of obtaining χ is: The probability of obtaining φ after realizing χ is: Thus, The probability of obtaining χ after measuring β is: Prob φ χ φ + Prob φ χ φ 79 Prob φ χ = χ φ 80 = χ χ + 3 χ 8 = 4 8 Prob χ φ = φ χ 83 = φ φ + 3 φ 84 = 4 85 Prob φ χ φ = 4/4/ = 6/69 86 Prob φ χ = χ φ 87 = χ χ + 3 χ 88 = 9 89

9 Logan A. Morrison Assignment Problem 6 Page 9 / 9 The probability of obtaining φ after realizing χ is: Therefore, Prob χ φ = φ χ 90 = 3 φ φ φ 9 = 9 9 Prob φ χ φ = 9/9/ = 8/69 93 Therefore, the probability of obtaining a after measuring β then α is Prob φ χ φ + Prob φ χ φ = =